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Sagot :
keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the equation above
[tex]x + y = 10\implies y = -x+10\implies y=\stackrel{\stackrel{m}{\downarrow }}{-1}x+10 \leftarrow \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}[/tex]
so, we're really looking for the equation of al ine whose slope is -1 and that passes through (-1 , 5)
[tex](\stackrel{x_1}{-1}~,~\stackrel{y_1}{5})\hspace{10em} \stackrel{slope}{m} ~=~ -1 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{5}=\stackrel{m}{-1}( ~~ x-\stackrel{x_1}{(-1) ~~ }) \\\\\\ y-5 = -(x+1)\implies y-5=-x-1\implies y=-x+4[/tex]
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